Problem: $ 27^{-\frac{4}{3}}$
Answer: $= \left(\dfrac{1}{27}\right)^{\frac{4}{3}}$ $= \left(\left(\dfrac{1}{27}\right)^{\frac{1}{3}}\right)^{4}$ To simplify $\left(\dfrac{1}{27}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left(? \right)^{3}=\dfrac{1}{27}$ To simplify $\left(\dfrac{1}{27}\right)^{\frac{1}{3}}$ , figure out what goes in the blank: $\left({\dfrac{1}{3}}\right)^{3}=\dfrac{1}{27}$ so $ \left(\dfrac{1}{27}\right)^{\frac{1}{3}}=\dfrac{1}{3}$ So $\left(\dfrac{1}{27}\right)^{\frac{4}{3}}=\left(\left(\dfrac{1}{27}\right)^{\frac{1}{3}}\right)^{4}=\left(\dfrac{1}{3}\right)^{4}$ $= \left(\dfrac{1}{3}\right)\cdot\left(\dfrac{1}{3}\right)\cdot \left(\dfrac{1}{3}\right)\cdot \left(\dfrac{1}{3}\right)$ $= \dfrac{1}{9}\cdot\left(\dfrac{1}{3}\right)\cdot \left(\dfrac{1}{3}\right)$ $= \dfrac{1}{27}\cdot\left(\dfrac{1}{3}\right)$ $= \dfrac{1}{81}$